What is the integral of #sec^3(x)#?

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Answer 1 · 2014-12-22T06:53:23

#I=int sec^3x dx#

by Integration by Pats with:
#u= secx# and #dv=sec^2x dx#
#=> du=secx tanx dx# and #v=tanx#,

#=secxtanx-int sec x tan^2x dx#

by #tan^2x=sec^2x-1#

#=secxtanx-int (sec^3x-secx) dx#

since #int sec^3xdx=I#,

#=secxtanx-I+int sec x dx#

by adding #I# and #int sec x dx=ln|secx+tanx|+C_1#

#=>2I=secxtanx+ln|secx+tanx|+C_1#

by dividing by 2,

#=>I=1/2secxtanx+1/2ln|secx+tanx|+C_1/2#

Hence,

#int sec^3 dx=1/2secxtanx+1/2ln|secx+tanx|+C#

I hope that this was helpful.